Proof of Theorem 1.1

Let G be a minimal counterexample for all the situations. Suppose that G has two distinct minimal normal subgroups $N_{1}$ and $N_{2}$ . Then, both $G/N_{1}$ and $G/N_{2}$ are solvable groups by induction. However, G is isomorphic to a subgroup of $G/N_{1}\times G/N_{2}$ , in contrast to the assumption that G is not solvable. Thus, G has a unique minimal normal subgroup in every situation of Theorem 1.1 and so it has at least one faithful irreducible character. Let N be the unique minimal normal subgroup of G. It follows that N is not solvable and $N'=N$ because $G/N$ is a solvable group by induction. Moreover, all faithful irreducible characters of G are strongly monolithic since $Z(G)=1$ .

Suppose that $\lvert \textrm {cd}_{\textrm {sm}}(G)\rvert \leq 2$ . Then $\lvert \textrm {cd}(G|N)\rvert \leq 2$ since $\textrm {cd}(G|N)\subseteq \textrm {cd}_{\textrm {sm}}(G)$ . It follows from [Reference Isaacs and Knutson8, Theorem B] that N is solvable. This contradiction completes the proof of (i).

Now, let $\nu \in \textrm {Irr}(N)$ and $\nu (1)>1$ . Then, there exists an irreducible character $\psi $ of G such that $[\psi ,\nu ^{G}]\ne 0$ . Thus, $\textrm {ker}\,\psi _{N}=N\cap \textrm {ker}\,\psi \leq \textrm {ker}\,\nu <N$ because $[\psi _{N},\nu ]\neq 0$ by Frobenius reciprocity. Hence, $\textrm {ker}\,\psi =1$ since N is the unique minimal normal subgroup of G and $\textrm {ker}\,\psi _{N}<N$ . Therefore, $\psi $ is a strongly monolithic character of G. Also, from Clifford’s theorem, $\nu (1)\mid \psi (1)$ . In the case (ii), $\nu (1)=\psi (1)$ and the degrees of all nonlinear irreducible characters of N are prime numbers. It follows from [Reference Isaacs and Passman9] that N is solvable, which is a contradiction. This completes the proof of (ii).

In the case (iii), N must be a proper subgroup of G. Otherwise, all nonlinear irreducible characters of G would be faithful and so strongly monolithic. However, this is a contradiction since all M-groups are solvable. Let $\chi $ be a faithful irreducible character of G of least degree. Then $\chi $ is a strongly monolithic character and so monomial by hypothesis. So there exists a linear character $\lambda $ of a subgroup H of G such that $\lambda ^{G}=\chi $ . Since $[G:H]=\chi (1)>1$ , it follows that $H<G$ and $(1_{H})^{G}$ is reducible. Let $\theta $ be an irreducible constituent of $(1_{H})^{G}$ , so that $\theta (1)<\chi (1)$ . By the choice of $\chi $ , we deduce that $N\leq \textrm {ker}\,\theta $ . Moreover, $N\leq \textrm {ker}\,(1_{H})^{G}\leq H$ because $N\leq \textrm {ker}\,\theta $ for every irreducible constituent $\theta $ of $(1_{H})^{G}$ . It follows that $N\leq \textrm {ker}\,\lambda $ since $N=N'$ and $\lambda $ is a linear character of H. Furthermore, $N\leq (\textrm {ker}\,\lambda )^{g}$ for every $g\in G$ , since $N\vartriangleleft G$ . This gives the contradiction that $N\leq \textrm {ker}(\lambda ^{G})=\textrm {ker}\chi =1$ . This contradiction completes the proof of (iii).

Proof of Theorem 1.2

If G is a nilpotent group, then by [Reference Gagola and Lewis4, Theorem A], $\chi (1)^{2}$ divides $\lvert G:\textrm {ker}\,\chi \rvert $ for every $\chi \in \textrm {Irr}(G)$ . Thus, it is clear that the necessary conditions in both cases of Theorem 1.2 hold.

Let G be a minimal counterexample for the sufficient conditions in both cases of the theorem. Since the hypotheses of Theorem 1.2 are inherited by $G/N$ for all $N\vartriangleleft G$ , it follows that G has a unique minimal normal subgroup M and therefore G has at least one faithful irreducible character. Moreover, $Z(G)=1=\Phi (G)$ since G is a minimal counterexample. Thus all faithful irreducible characters of G are strongly monolithic. By the hypothesis of case (i) in the theorem, $\theta (1)^{2}\mid \lvert G:\textrm {ker}\,\theta \rvert $ for every faithful irreducible character $\theta $ of G. Moreover, $\chi (1)^{2}\mid \lvert G:\textrm {ker}\,\chi \rvert $ for every $\chi \in \textrm {Irr}(G)$ , because $G/M$ is nilpotent by induction. Thus, by [Reference Gagola and Lewis4, Theorem A], G is nilpotent, which is a contradiction. This contradiction completes the proof of (i).

Since $\Phi (G)=1$ , there exists a subgroup H of G such that $G=MH$ and $M\cap H=1$ . Then H is nilpotent by induction. Using Gaschütz’s theorem, $(\lvert M\rvert ,\lvert H\rvert )=1$ . It follows from [Reference Isaacs6, Theorem B] that there exists a $\lambda \in \textrm {Irr}(M)$ with $\lambda \neq 1$ such that $\lvert I_{H}(\lambda )\rvert \leq (\lvert {H\rvert }/{q})^{1/q}$ , where $I_{H}(\lambda )$ is the inertia group of $\lambda $ in H and q is the smallest prime divisor of $\lvert H\rvert $ . This gives the inequality:

$$ \begin{align*} \lvert I_{H}(\lambda)\rvert \leq\bigg(\frac{\lvert H\rvert}{q}\bigg)^{1/q}\leq\bigg(\frac{\lvert H\rvert}{2}\bigg)^{1/2}<\lvert H\rvert^{1/2}. \end{align*} $$

From [Reference Isaacs5, Problem 6.18], there exists a linear character $\psi \in \textrm {Irr}(I_{G}(\lambda ))$ such that $\psi _{M} = \lambda $ , where $I_{G}(\lambda )$ is the inertia group of $\lambda $ in G. This implies that $\psi ^{G}$ is a monomial irreducible character of G. Let $\chi =\psi ^{G}$ . Then $\chi $ is a faithful irreducible character of G. Otherwise, $M\leq \textrm {ker}\,\chi ={\bigcap }_{g\in G}(\textrm {ker}\,\psi )^{g}\leq \textrm {ker}\,\psi $ . However, this is a contradiction since $\psi _{M}=\lambda \neq 1$ . Therefore, $\chi $ is a monomial strongly monolithic character of G. By the hypothesis of (ii), $\chi (1)^{2}\mid \lvert G\rvert $ . Also $\chi (1)\mid \lvert H\rvert $ by Ito’s theorem (see [Reference Isaacs5]). Then, $\lvert G\rvert =\lvert M\rvert \lvert I_{H}(\lambda )\rvert\kern2pt \chi (1)<\lvert M\rvert \lvert H\rvert =\lvert G\rvert $ , which is a contradiction. This completes the proof of (ii).

Proof of Theorem 1.3

Let G be a minimal counterexample to the case (i). Assume that G has two distinct minimal normal subgroups, say $N_{1}$ and $N_{2}$ . It follows that $\lvert \textrm {ker}_{\textrm {msm}}(G/N_{i})\rvert \leq 1$ for $i=1,2$ . Since a solvable group does not have any monomial strongly monolithic characters if and only if it is abelian, the commutator subgroups of $G/N_{1}$ and $G/N_{2}$ are nilpotent by induction. Thus, $G'$ is nilpotent since G is isomorphic to a subgroup of $G/N_{1}\times G/N_{2}$ . This contradiction shows that G has a unique minimal normal subgroup and it has at least one faithful irreducible character.

Let N be the unique minimal normal subgroup of G. Assume that $Z(G)>1$ . Then $G'Z(G)/Z(G)$ , the commutator subgroup of $G/Z(G)$ , is nilpotent by induction. Therefore, $G'$ is also nilpotent, which is a contradiction. Thus, $Z(G)$ must be trivial and so all faithful irreducible characters of G must be strongly monolithic. However, $\Phi (G)=1$ . Otherwise, $G'\Phi (G)/\Phi (G)$ would be nilpotent by induction. Thus, $G'\Phi (G)$ would be nilpotent by [Reference Isaacs7, Theorem 8.24], which is a contradiction. Since $\Phi (G)=1$ , there is a subgroup H of G such that $G=NH$ and $N\cap H=1$ . Let $1_{N}\neq \lambda \in \textrm {Irr}(N)$ . From [Reference Isaacs5, Problem 6.18], there exists a linear character $\theta \in \textrm {Irr}(I_{G}(\lambda ))$ such that $\theta _{N} = \lambda $ , where $I_{G}(\lambda )$ is the inertia group of $\lambda $ in G. This implies that $\theta ^{G}\in \textrm {Irr}(G)$ is a faithful monomial strongly monolithic character of G since $\textrm {ker}\,\theta ^{G}=1$ . Since $\lvert \textrm {ker}_{\textrm {msm}}(G)\rvert =1$ , the maximal kernel among the kernels of all nonlinear irreducible characters of G must be trivial. It follows that all nonlinear irreducible characters of G are faithful. Thus, $G'$ is the unique minimal normal subgroup of G, which is a contradiction. This completes the proof of (i).

To prove (ii), let G again be a minimal counterexample. Then G has a unique minimal normal subgroup N. Assume that $Z(G)>1$ . Then $G'Z(G)/Z(G)$ is meta-nilpotent by induction. Thus, there is a normal subgroup A of $G'Z(G)$ such that both $G'Z(G)/A$ and $A/Z(G)$ are nilpotent. Since $A/Z(G)$ is nilpotent, A is nilpotent and so $A\cap G'$ is nilpotent. Moreover, $G'/(A\cap G')$ is nilpotent because $G'/(A\cap G')\cong G'Z(G)/A$ . Then $G'$ is also meta-nilpotent, which is a contradiction. Hence, $Z(G)=1$ . Similarly, it is clear that $\Phi (G)=1$ . Therefore, G has a faithful monomial strongly monolithic character $\chi $ as in the proof of (i). Since $\lvert \textrm {ker}_{\textrm {msm}}(G)\rvert =2$ , we see that $\lvert \textrm {ker}_{\textrm {msm}}(G/N)\rvert =1$ . Thus, $G'/N$ is nilpotent by (i) of the theorem, that is, $G'$ is meta-nilpotent. This contradiction completes the proof of (ii).